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Hidden Markov Model

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Hidden Markov Model. CS570 Lecture Note KAIST. This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB). Sequential Data. Often highly variable, but has an embedded structure Information is contained in the structure. More examples. - PowerPoint PPT Presentation
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1 This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Pro f. Wilensky (UCB) Hidden Markov Model CS570 Lecture Note KAIST This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
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Hidden Markov Model TutorialThis lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Hidden Markov Model
CS570 Lecture Note
This lecture note was made based on the notes of
Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Sequential Data
Information is contained in the structure
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
More examples
main() { char q=34, n=10, *a=“main() {
char q=34, n=10, *a=%c%s%c; printf(
a,q,a,q,n);}%c”; printf(a,q,a,n); }
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Example: Speech Recognition
Given a sequence of inputs-features of some kind extracted by some hardware, guess the words to which the features correspond.
Hard because features dependent on
Speaker, speed, noise, nearby features(“co-articulation” constraints), word boundaries
“How to wreak a nice beach.”
“How to recognize speech.”
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Defining the problem
y, a string of acoustic features of some form,
w, a string of words, from some fixed vocabulary
L, a language (defined as all possible strings in that language),
Given some features, what is the most probable string the speaker uttered?
Pukyong Natl Univ
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Analysis
P(w|y)= P(w)P(y|w)/P(y)
Since y is the same for different w’s we might choose, the
problem reduces to
each possible string in our language
pronunciation, given an utterance.
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P(w) where w is an utterance
Problem: There are a very large number of possible utterances!
Indeed, we create new utterances all the time, so we cannot hope to have there probabilities.
So, we will need to make some independence assumptions.
First attempt: Assume that words are uttered independently of one another.
Then P(w) becomes P(w1)… P(wn) , where wi are the individual words in the string.
Easy to estimate these numbers-count the relative frequency words in the language.
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Assumptions
Words don’t just follow each other randomly
Second attempt: Assume each word depends only on the previous word.
E.g., “the” is more likely to be followed by “ball” than by “a”,
despite the fact that “a” would otherwise be a very common, and hence, highly probably word.
Of course, this is still not a great assumption, but it may be a decent approximation
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
In General
This is typical of lots of problems, in which
we view the probability of some event as dependent on potentially many past events,
of which there too many actual dependencies to deal with.
So we simplify by making assumption that
Each event depends only on previous event, and
it doesn’t make any difference when these events happen
in the sequence.
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Speech Example
= s p iy iy iy ch ch ch ch
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Analysis Methods
Probability-based analysis?
Method I
A poor model for temporal structure
Model size = |V| = N
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Analysis methods
Method II
A symbol is dependent only on the immediately preceding:
|V|×|V| matrix model
105×105 – doubly outrageous!!
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Another analysis method
Method III
What you see is a clue to what lies behind and is not known a priori
The source that generated the observation
The source evolves and generates characteristic observation sequences
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More Formally
To clarify, let’s write the sequence this way:
P(q1=Si, q2=Sj,…, qn-1=Sk, qn=Si)
Here the indicate the I-th position of the sequence,
and the Si the possible different words from our
vocabulary.
E.g., if the string were “The girl saw the boy”, we might have
S1= the q1= S1
S2= girl q2= S2
S3= saw q3= S3
S4= boy q4= S1
S1= the q5= S4
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Formalization (continue)
We want P(q1=Si, q2=Sj,…, qn-1=Sk, qn=Si)
Let’s break this down as we usually break down a joint :
= P(qn=Si | q1=Sj,…,qn-1=Sk)P(q1=Sj,…,qn-1=Sk)

qn-1=Sm)P(q2=Sj|q1=Sj)P(q1=Si)
Our simplifying assumption is that each event is only
dependent on the previous event, and that we don’t care when
the events happen, I.e.,
P(qi=Si | qi-1=Sk)=P(qj=Si | qj-1=Sk)
This is called the Markov assumption.
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Markov Assumption
“The future does not depend on the past, given the present.”
Sometimes this if called the first-order Markov assumption.
second-order assumption would mean that each event depends on the previous two events.
This isn’t really a crucial distinction.
What’s crucial is that there is some limit on how far we are willing to look back.
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Morkov Models
The Markov assumption means that there is only one probability to remember for each event type (e.g., word) to another event type.
Plus the probabilities of starting with a particular event.
This lets us define a Markov model as:
finite state automaton in which
the states represent possible event types(e.g., the different words in our example)
the transitions represent the probability of one event type following another.
It is easy to depict a Markov model as a graph.
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Example: A Markov Model for a Tiny Fragment
of English
e.g., P(qi=girl|qi-1 =the)=.8
The numbers on the initial arrows show the probability of
starting in the given state.
Missing probabilities are assumed to be O.
the
a
girl
little
.7
.8
.9
.22
.3
.78
.2
.1
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Example: A Markov Model for a Tiny Fragment
of English
with probabilities :
P(“A little little girl”)=.3.22.1.9= .00594
the
a
girl
little
.7
.8
.9
.22
.3
.78
.2
.1
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Example(con’t)
P(q1=the, q2=little, q3=girl)
where , and are states.
“Given that sentence begins with “a”, what is the probability
that the next words were “little girl”?”
P(q3=the, q2=little, q1=a)
= P(q3=girl | q2=little, q1=a)P(q2=little q1=a)
= P(q3=girl | q2=little)P(q2=little q1=a)
= .9.22=.198
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Markov Models and Graphical Models
Markov models and Belief Networks can both be represented by nice graphs.
Do the graphs mean the same thing?
No! In the graphs for Markov models; nodes do not represent random variables, CPTs.
Suppose we wanted to encode the same information via a belief network.
We would have to “unroll” it into a sequence of nodes-as many as there are elements in the sequence-each dependent on the previous, each with the same CPT.
This redrawing is valid, and sometimes useful, but doesn’t explicitly represent useful facts, such as that the CPTs are the same everywhere.
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Back to the Speech Recognition Problem
A Markov model for all of English would have one node for each word, which would be connected to the node for each word that can follow it.
Without Loss Of Generality, we could have connections from every node to every node, some of which have transition probability 0.
Such a model is sometimes called a bigram model.
This is equivalent to knowing the probability distribution of pair of words in sequences (and the probability distribution for individual words).
A bigram model is an example of a language model, i.e., some (in this case, extremely simply) view of what sentences or sequences are likely to be seen.
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Bigram Model
Bigram models are rather inaccurate language models.
E.g., the word after “a” is much more likely to be “missile” if the word preceding “a” is “launch”.
the Markov assumption is pretty bad.
If we could condition on a few previous words, life gets a bit better:
E.g., we could predict “missile” is more likely to follow “launch a” than “saw a”.
This would require a “second order” Markov model.
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Higher-Order Models
In the case of words, this is equivalent to going to trigrams.
Fundamentally, this isn’t a big difference:
We can convert a second order model into a first order model, but with a lot more states.
And we would need much more data!
Note, though, that a second-order model still couldn’t accurately predict what follows “launch a large”
i.e., we are predicting the next work based on only the two previous words, so the useful information before “a large” is lost.
Nevertheless, such language models are very useful approximations.
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Back to Our Spoken Sentence recognition Problem
We are trying to find
argmaxw∈L P(w)P(y|w)
Now let’s look at P(y|w).
- That is, how do we pronounce a sequence of words?
- Can make the simplification that how we pronounce words
is independent of one another.
P(y|w)=ΣP(o1=vi,o2=vj,…,ok=vl|w1)×
… × P(ox-m=vp,ox-m+1=vq,…,ox=vr |wn)
i.e., each word produces some of the sounds with some
probability; we have to sum over possible different word
boundaries.
So, what we need is model of how we pronounce individual words.
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A Model
We can represent this idea by complicating the Markov model:
- Let’s add probabilistic emissions of outputs from each state.
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Example: A (Simplistic) Model for Pronouncing “of”
Each state can emit a different sound, with some probability.
Variant: Have the emissions on the transitions, rather than
the states.
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How the Model Works
We can’t “see” the actual state transitions.
But we can infer possible underlying transitions from the
observations, and then assign a probability to them
E.g., from “o v”, we infer the transition “phone1 phone2”
- with probability .7 x .9 = .63.
I.e., the probability that the word “of” would be pronounced
as “o v” is 63%.
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Hidden Markov Models
This is a “hidden Markov model”, or HMM.
Like (fully observable) Markov models, transitions from one
state to another are independent of everything else.
Also, the emission of an output from a state depends only on
that state, i.e.:
=P(o1|q1)×P(o2|q2)×…×P(on|q1)
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HMMs Assign Probabilities
We want to know how probable a sequence of observations
is given an HMM.
ways to produce the observed output.
So, we have to consider all possible ways an output might be
produced, i.e., for a given HMM:
P(O) = ∑Q P(O|Q)P(Q)
where O is a given output sequence, and Q ranges over all
possible sequence of states in the model.
P(Q) is computed as for (visible) Markov models.
P(O|Q) = P(o1,o2,…,on|q1,…qn)
= P(o1|q1)×P(o2|q2)×…×P(on|q1)
We’ll look at computing this efficiently in a while…
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Finishing Solving the Speech Problem
To find argmaxw∈L P(w)P(y|w), just consider these
probabilities over all possible strings of words.
Could “splice in” each word in language model with its HMM
pronunciation model to get one big HMM.
- Lets us incorporate more dependencies.
- E.g., could have two models of “of”, one of which has a
much higher probability of transitioning to words beginning
with consonants.
are broken up into acoustic vectors.
- but these are also HMMs.
- So we can make one gigantic HMM out of the whole thing.
So, given y, all we need is to find most probable path through
the model that generates it.
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Example: Our Word Markov Model
the
a
girl
little
.7
.8
.9
.22
.3
.78
.2
.1
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Example: Splicing in Pronunciation HMMs
V1
V2
V3
V4
V5
V6
V7
V8
V9
V10
V11
V12
V9
V10
V8
V11
V12
.8
.7
.2
.78
.3
.22
.1
a
.9
little
girl
the
1
2
3
4
5
6
8
9
10
7
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Example: Best Sequence
girl
V1
V2
V3
V4
V5
V6
V7
V8
V9
V10
V11
V12
V9
V10
V8
V11
V12
.8
.7
.2
.78
.3
.22
.1
a
.9
little
the
Suppose observation is “v1 v3 v4 v9 v8 v11 v7 v8 v10”
Suppose most probable sequence is determined to be
“1,2,3,8,9,10,4,5,6” (happens to be only way in example)
Then interpretation is “the little girl”.
1
2
3
4
5
6
8
9
10
7
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Hidden Markov Models
Determine the probability of a give sequence
Determine the probability of a model producing a sequence in a particular way
equivalent to recognizing or interpreting that sequence
Learning a model form some observations.
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Why HMM?
Because the HMM is a very good model for such patterns!
highly variable spatiotemporal data sequence
often unclear, uncertain, and incomplete
Because it is very successful in many applications!
Because it is quite easy to use!
Tools already exist…
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The problem
Not quite a valid assumption
There are often errors or noise
Noisy sound, sloppy handwriting, ungrammatical or Kornglish sentence
There may be some truth process
Underlying hidden sequence
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The Auxiliary Variable
nor is
is Markovian
“Markov chain”
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Summary of the Concept
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Hidden Markov Model
stochastic chain process : { q(t) }
output process : { f(x|q) }
is also called as
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HMM Characterization
(A, B, )
B : symbol output/observation probability
: initial state distribution probability
{ i | i = p(q1=i) }
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HMM, Formally
qt denotes the state at time t.
A transition probability matrix A, such that
A[i,j]=aij=P(qt+1=Sj|qt=Si)
This is an N x N matrix.
A set of symbols, {v1,…,vM}
For all purposes, these might as well just be {1,…,M}
ot denotes the observation at time t.
A observation symbol probability distribution matrix B, such that
B[i,j]=bi,j=P(ot=vj|qt=Si)
This is a N x M matrix.
An initial state distribution, π, such that πi=P(q1=Si)
For convenience, call the entire model λ = (A,B,π)
Note that N and M are important implicit parameters.
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Graphical Example
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Data interpretation
P(s s p p iy iy iy ch ch ch|)
= Q P(ssppiyiyiychchch,Q|)
P(Q|) p(ssppiyiyiychchch|Q, )
= P(1122333444|) p(ssppiyiyiychchch|1122333444, )
= P(1| )P(s|1,) P(1|1, )P(s|1,) P(2|1, )P(p|2,)
P(2|2, )P(p|2,) …..
×(.3×.6)×(1.×.6)2
0.6 0.4 0.0 0.0
0.0 0.5 0.5 0.0
0.0 0.0 0.7 0.3
0.0 0.0 0.0 1.0
0.2 0.2 0.0 0.6 …
0.0 0.2 0.5 0.3 …
0.0 0.8 0.1 0.1 …
0.6 0.0 0.2 0.2 …
Let Q = 1 1 2 2 3 3 3 4 4 4
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Issues in HMM
Difficult problems
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The Number of States
r r g b b g b b b r
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(1) The simplest model
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(2) Two state model
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(3) Three state models
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The Criterion is
The best topology comes from insight and experience
the # classes/symbols/samples
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A trained HMM
.5 .4 .1
.0 .6 .4
.0 .0 .0
.6 .2 .2
.2 .5 .3
.0 .3 .7
1. 0. 0.
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Three Problems
What is the probability of the observation?
Given an observed sequence and an HMM, how probable is that sequence?
Forward algorithm
What is the best state sequence for the observation?
Given an observed sequence and an HMM, what is the most likely state sequence that generated it?
Viterbi algorithm
Given an observation, can we learn an HMM for it?
Baum-Welch reestimation algorithm
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Problem: How to Compute P(O|λ)
P(O|λ)=∑all Q P(O|Q,λ)P(Q|λ)
where O is an observation sequence and Q is a sequence of
states.
the model, and sum up their probabilities.
Naively, this would involve O(TNT) operations:
- At every step, there are N possible transitions, and there are
T steps.
However, we can take advantage of the fact that, at any given
time, we can only be in one of N states, so we only have to
keep track of paths to each state.
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Basic Idea for P(O|λ) Algorithm
If we know the probability
of being in each state at
time t, and producing the
observation so far (αt(i)),
then the probability of
next clock tick, and then
emitting the next output, is
easy to compute.
time t+1
αt(i)
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A Better Algorithm For Computing P(O|λ)
Let αt(i) be defined as P(o1o2…ot,qt=Si|λ).
- I.e., the probability of seeing a prefix of the observation, and
ending in a particular state.
Algorithm:
- Induction: αt+1(j) ← (∑1≤i≤N αt(i)aij)bj,ot+1
1 ≤ t ≤ T-1, 1 ≤ j ≤ N
- Finish: Return ∑1≤i≤N αT(i)
This is called the forward procedure.
How efficient is it?
- At each time step, do O(N) multiplications and additions for
each of N nodes, or O(N2T).
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Can We Do it Backwards?
βt(i) is probability that,
If we know probability of
outputting the tail of the
observation, given that we
compute probability that,
at the previous clock tick,
we emit the (one symbol
bigger) tail.
βt+1(i)
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Going Backwards Instead
Define βt(i) as P(ot+1ot+2…oT|qt=Si,λ).
- I.e., the probability of seeing a tail of the observation, given
that we were in a particular state.
Algorithm:
- Induction: βt(i) ← ∑1≤j≤N aijbj,ot+1βt+1(j)
T-1 ≥ t ≥ 1, 1 ≤ i ≤ N
This is called the backward procedure.
We could use this to compute P(O|λ) too (β1(i) is
P(o2o3…oT|q1=Si,λ), so P(O|λ)= ∑1≤j≤Nπibi,o1β1(i)).
- But nobody does this.
How efficient is it?
- Same as forward procedure.
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1. Model Evaluation
Solution: forward/backward procedure
Define: forward probability -> FW procedure
Define: backward probability -> BW procedure
These are probabilities of the partial events leading to/from a point in space-time
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Forward procedure
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Numerical example: P(RRGB|)
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Backward procedure
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2nd Problem: What’s the Most Likely State Sequence?
Actually, there is more than one possible interpretation of
“most likely state sequence”.
One is, which states are individually most likely.
- i.e., what is the most likely first state? The most likely second?
And so on.
- Note, though, that we can end up with a “sequence” that isn’t
even a possible sequence through the HMM, much less a
likely one.
- i.e., find argmaxQ P(Q|O,λ)
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Algorithm Idea
Suppose we knew the highest probability path ending in each
state at time step t.
We can compute the highest probability path ending in a state
at t+1 by considering each transition from each state at time t,
and remembering only the best one.
This is another application of the dynamic programming
principle.
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Basic Idea for argmaxQP(Q|O,λ) Algorithm
If we know the probability
of the best path to each
state at time t producing
the observation so far
(δt(i)), then the probability
state producing the next
observation at the next
probability times the
transition probability, times
emission.
a1,4
a2,4
aN-1,4
aN,4
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Definitions for Computing argmaxQ P(Q|O,λ)
Note that P(Q|O,λ)=P(Q,O|λ)/P(O|λ), so maximizing
P(Q|O,λ) is equivalent to maximizing P(Q,O|λ).
- Turns out latter is a bit easier to work with.
Define δt(i) as
maxq1,q2,…,qt-1P(q1,q2,…,qt=i,o1,o2,…,ot|λ).
We’ll use these to inductively find P(Q,O|λ). But how do we
find the actual path?
We’ll maintain another array, ψt(i), which keeps track of the
argument that maximizes δt(i).
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Algorithm for Computing argmaxQ P(Q|O,λ)
Algorithm:
ψ1(i) ← 0
- Finish: P* = maxi (δT(i)) is probability of best path
qT* = argmaxi (δT(i)) is best final state
- Extract path by backtracking:
This is called the Viterbi algorithm.
Note that it is very similar to the forward procedure (and hence
as efficient).
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
2. Decoding Problem
The best path Q* given an input X ?
It can be obtained by maximizing the joint probability over state sequences
Path likelihood score:
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Viterbi algorithm
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Numerical Example: P(RRGB,Q*|)
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
3. Model Training Problem
No analytical solution exists
MLE + EM algorithm developed
Baum-Welch reestimation [Baum+68,70]
maximizes the probability estimate of observed events
guarantees finite improvement
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
MLE Example
Experiment
Known: 3 balls inside (some white, some red; exact numbers unknown)
Unknown: R = # red balls
Two models
p (|R=3) = 3C2 / 3C2 = 1
Which model?
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Learning an HMM
We will assume we have an observation O, and want to know
the “best” HMM for it.
I.e., we would want to find argmaxλ P(λ|O), where
λ= (A,B,π) is some HMM.
- I.e., what model is most likely, given the observation?
When we have a fixed observation that we use to pick a model,
we call the observation training data.
Functions/values like P(λ|O) are called likelihoods.
What we want is the maximum likelihood estimate(MLE) of the
parameters of our model, given some training data.
This is an estimate of the parameters, because it depends for
its accuracy on how representative the observation is.
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Maximizing Likelihoods
- The observation is constant, so it is enough to maximize
P(λ)P(O|λ).
P(λ) is the prior probability of a model; P(O|λ) is the
probability of the observation, given a model.
Typically, we don’t know much about P(λ).
- E.g., we might assume all models are equally likely.
- Or, we might stipulate that some subclass are equally likely,
and the rest not worth considering.
We will ignore P(λ), and simply optimize P(O|λ).
I.e., we can find the most probable model by asking which
model makes the data most probable.
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Maximum Likelihood Example
Simple example: We have a coin that may be biased. We
would like to know the probability that it will come up heads.
Let’s flip it a number of times; use percentage of times it comes
up heads to estimate the desired probability.
Given m out of n trials come up heads, what probability should
be assigned?
In terms of likelihoods, we want to know P(λ|O), where our
model is just the simple parameter, the probability of a coin
coming up heads.
maximizing P(λ)P(O|λ).
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Maximizing P(λ)P(O|λ) For Coin Flips
Note that knowing something about P(λ) is knowing whether
coins tended to be biased.
If we don’t, we just optimize P(O|λ).
I.e., let’s pick the model that makes the observation most likely.
We can solve this analytically:
- P(m heads over n coin tosses|P(Heads)=p) = nCmpm(1-p)n-m
- Take derivative, set equal to 0, solve for p.
Turns out p is m/n.
- This is probably what you guessed.
So we have made a simple maximum likelihood estimate.
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Comments
Clearly, our result will just be an estimate.
- but one we hope will become increasingly accurate with
more data.
Note, though, via MLE, the probability of everything we haven’t
seen so far is 0.
For modeling rare events, there will never be enough data for
this problem to go away.
There are ways to smooth over this problem (in fact, one way
is called “smoothing”!), but we won’t worry about this now.
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Back to HMMs
MLE tells us to optimize P(O|λ).
We know how to compute P(O|λ) for a given model.
- E.g., use the “forward” procedure.
How do we find the best one?
Unfortunately, we don’t know how to solve this problem
analytically.
However, there is a procedure to find a better model, given an
existing one.
- So, if we have a good guess, we can make it better.
- Or, start out with fully connected HMM (of given N, M) in
which each state can emit every possible value; set all
probabilities to random non-zero values.
Will this guarantee us a best solution?
No! So this is a form of …
- Hillclimbing!
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Maximizing the Probability of the Observation
Basic idea is:
2. Compute new λnew based on λold and observation O.
3. If P(O|λnew)-P(O| λold) < threshold
(or we’ve iterated enough), stop.
4. Otherwise, λold ← λnew and go to step 2.
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Increasing the Probability of the Observation
Let’s “count” the number of times, from each state, we
- start in a given state
- make a transition to each other state
- emit each different symbol.
If we knew these numbers, we can compute new probability
estimates.
We can’t really “count” these.
- We don’t know for sure which path through the model was
taken.
- But we know the probability of each path, given the model.
So we can compute the expected value of each figure.
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Set Up
- i.e., the probability that, given observation and model, we
are in state Si at time t and state Sj at time t+1.
Here is how such a transition can happen:
Si
Sj
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Computing ξt(i,j)
But we want ξt(i,j) = P(qt=Si,qt+1=Sj|O,λ)
By definition of conditional probability,
ξt(i,j) = αt(i)aijbj,ot+1βt+1(j)/P(O|λ)
i.e., given
- P(O), which we can compute by forward, but also, by
∑i ∑j αt(i)aijbj,ot+1βt+1(j)
- and A,B, which are given in model, – we can compute ξ.
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
One more preliminary…
Define γt(i) as probability of being in state Si at time t, given
the observation and the model.
Given our definition of ξ:
γt(i)=∑1≤j≤N ξt(i,j)
So:
- ∑1≤t≤T-1 γt(i) = expected number of transitions from Si
- ∑1≤t≤T-1 ξt(i,j) = expected number of transitions from
Si to Sj
Pukyong Natl Univ
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Now We Can Reestimate Parameters
aij’ = expected no. of transitions from Si to Sj/expected no. of
transitions from Si
πi’ = probability of being in S1 = γ1(i)
bjk’ = expected no. of times in Sj, observing vk/expected no.
of times in Sj
= ∑1≤t≤T, s.t. Ot=vkγt(i)/∑1≤t≤T γt(i)
Pukyong Natl Univ
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Iterative Reestimation Formulae
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Baum-Welch Algorithm
or Forward-backward algorithm
It was proven (by Baum et al.) that this reestimation procedure
leads to increased likelihood.
But remember, it only guarantees climbing to a local maximum!
It is a special case of a very general algorithm for incremental
improvement by iteratively
transitions and state emissions),
- The general procedure is called Expectation-Maximization,
or the EM algorithm.
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Implementation Considerations
observation, and eventually underflow.
- I.e., instead of computing αt(i), at each time step, compute
αt(i)/∑iαt(i).
- Turns out that if you also used the ∑iαt(i)s to scale the
βt(j)s, you get a numerically nice value, although it doesn’t
has a nice probabilities interpretation;
- Yet, when you use both scaled values, the rest of the ξt(i,j)
computation is exactly the same.
Another: Sometimes estimated probabilities will still get very
small, and it seems better to not let these fall all the way to 0.
Pukyong Natl Univ
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Other issues
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Other types of parametric structure
Continuous density HMM (CHMM)
Semi-continuous HMM
State-duration HMM
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Graphical DHMM and CHMM
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Statistical Decision Making System
Statistical K-class pattern recognition
1, …, K
P(X|k)
P(k)
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An Application of HMMs:“Part of Speech” Tagging
A problem: Word sense disambiguation.
- I.e., words typically have multiple senses; need to determine
which one a given word is being used as.
For now, we just want to guess the “part of speech” of each
word in a sentence.
function of their “grammatical class”.
- Grammatical classes are things like verb, noun, preposition,
determiner, etc.
- Words often have multiple grammatical classes.
» For example, the word “rock” can be a noun or a verb.
» Each of which can have a number of different meanings.
We want an algorithm that will “tag” each word with its most
likely part of speech.
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Parsing, Briefly
S
NP
VP
V
PRO
NP
D
N
I
saw
a
bird
Pukyong Natl Univ
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However
Each of these words is listed in the dictionary as having
multiple POS entries:
- “saw”: noun, verb
birdwatch”)
- “I”: pronoun, noun (the letter “I”, the square root of –1,
something shaped like an I (I-beam), symbol (I-80, Roman
numeral, iodine)
- “a”: article, noun (the letter “a”, something shaped like an “a”,
the grade “A”), preposition (“three times a day”), French
pronoun.
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Moreover, there is a parse!
S
NP
VP
N
N
N
V
I
saw
a
bird
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
What’s It Mean?
The question is, can we avoid such nonsense cheaply.
Note that this parse corresponds to a very unlikely set of POS
tags.
So, just restricting our tags to reasonably probably ones might
eliminate such silly options.
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
Solving the Problem
- Using “tagged corpora”.
How well does it work?
Turns out it will be correct about 91% of the time.
Good?
Humans will agree with each other about 97-98% of the time.
So, room for improvement.
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
A Better Algorithm
Make POS guess depend on context.
For example, if the previous word were “the”, then the word
“rock” is much more likely to be occurring as a noun than as
a verb.
We can incorporate context by setting up an HMM in which the
hidden states are POSs, and the words the emissions in those
states.
Pukyong Natl Univ
This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
HMM For POS Tagging:Example
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This lecture note was made based on the notes of Prof. B.K.Shin(Pukyung Nat’l Univ) and Prof. Wilensky (UCB)
HMM For POS Tagging
Second-order equivalent to POS trigrams.
Generally works well if there is a hand-tagged corpus from
which to read off the probabilities.
- Lots of detailed issues: smoothing, etc.
If none available, train using Baum-Welch.
Usually start with some constraints:
- E.g., start with 0 emissions for words not listed in dictionary
as having a given POS; estimate transitions.
Best variations get about 96-97% accuracy, which is
approaching human performance.
Pukyong Natl Univ

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